Firstly before we begin, your plane should be diving 30 degrees downwards. It is important that we are clear on this as this means the bomb would have an initial downwards vertical velocity when it is released, i.e. The bomb is already travelling downwards instead of being stationary when released. Let this velocity be v ms^-1.
So now, let us start by looking at the forces involved. At the point when the bomb is released, only gravity is acting on the bomb, thus assuming the bomb does not experience air resistance, it accelerates downwards at g, which for the sake of convenience I would use g=10ms^-2 for this explanation (do use g=9.8 or 9.82 depending on what value of g is expected). The bomb does not experience any horizontal force, hence no horizontal acceleration.
Now, the bomb covers 1000 meters of height in 10 seconds. Constructing a formula for the distance covered, we get 10v+0.5*g*10^2=1000, which is v=50ms^-1.
This is not the answer yet, as this value is just the vertical component of the plane. Calculating the velocity of the plane is now just simple trigonometry, with velocity of the plane being 50sin(30), which brings us to the final answer of 100ms^-1.
(sorry couldn't give you a graphical explanation as I'm still outside being bored >< any questions just ask Kays jyy :)))).)
Edit: you might have subbed in t=cos 30 by mistake, t is 10 and you might want to try it again